Question

the numerator of a fraction is 3 less than the denominator. If 4 is added to both the numerator and denominator, the value of fraction increases by 1/8 , find the fraction.

Answer 1

Answer:

If x=8 then fraction becomes

$\frac{8-3}{8}=\frac{5}{8}$

If x=-12 then fraction becomes

$\frac{-12-3}{-12}=\frac{-15}{-12}$  

Step-by-step explanation:

Given that the numerator of a fraction is 3 less than the denominator. If 4 is added to both the numerator and denominator, the value of fraction increases by 1/8 ,

we have to find the fraction.

Let the denominator of fraction is x

As numerator of a fraction is 3 less than the denominator

∴ numerator is x-3

$\text{The fraction becomes }\frac{x-3}{x}$

Now, if 4 is added to both the numerator and denominator, the value of fraction increases by 1/8 ,

$\frac{x-3+4}{x+4}=\frac{x-3}{x}+\frac{1}{8}$

$\frac{x+1}{x+4}-\frac{x-3}{x}=\frac{1}{8}$

$x(x+1)-(x+4)(x-3)=\frac{1}{8}x(x+4)$

$x^2+4x-96=0$

$(x-8)(x+12)=0$

if x=8 then fraction becomes

$\frac{8-3}{8}=\frac{5}{8}$

If x=-12 then fraction becomes

$\frac{-12-3}{-12}=\frac{-15}{-12}$

Answer 2

Answer:

If x=8 then fraction becomes

\frac{8-3}{8}=\frac{5}{8}

8

8−3

=

8

5

If x=-12 then fraction becomes

\frac{-12-3}{-12}=\frac{-15}{-12}

−12

−12−3

=

−12

−15

Step-by-step explanation:

Given that the numerator of a fraction is 3 less than the denominator. If 4 is added to both the numerator and denominator, the value of fraction increases by 1/8 ,

we have to find the fraction.

Let the denominator of fraction is x

As numerator of a fraction is 3 less than the denominator

∴ numerator is x-3

\text{The fraction becomes }\frac{x-3}{x}The fraction becomes

x

x−3